# Scalar field

1. Sep 29, 2013

### PhysicsGente

1. The problem statement, all variables and given/known data

This is not really a problem but I was going over my lecture notes and I see $\mathscr{H}=\frac{1}{2}\left(\pi^{2} + \vec{\nabla}\phi \cdot \vec{\nabla}\phi + m^{2}\phi^{2}\right)$ and $\frac{\partial\mathscr{H}}{\partial\phi} = -\nabla^{2}\phi + m^{2}\phi$

2. Relevant equations

3. The attempt at a solution

I would think that $\frac{\partial\mathscr{H}}{\partial\phi} = \nabla^{2}\phi + m^{2}\phi$. But I don't know where the minus sign is coming from.

2. Sep 29, 2013

### PhysicsGente

I just found a vector identity $\vec{\nabla}\phi\cdot\vec{\nabla}\phi = \vec{\nabla}\cdot\left(\phi\vec{\nabla}\phi\right) - \phi\vec{\nabla}^{2}\phi$. I now see how the result follow.

EDIT: I'm confused again. will the phi derivative of the first term vanish?

3. Sep 30, 2013

### Sonny Liston

If by "the first term" you mean the $\pi^2$ term, then yes, the $\phi$ derivative will kill that term; reason being $\phi$ doesn't appear in that term, only its time derivative.

Edit: Sorry, I understand what you meant by "first term" now. The first term vanishes at infinity so you can ignore it.

Last edited: Sep 30, 2013
4. Sep 30, 2013

### dextercioby

The 3-divergence in the Hamiltonian density (i.e. the first term in the RHS of the equality in post #2) can be discarded, since by integration of full space gives 0, because scalar fields are normally taken from the Schwartz space of test functions.

5. Oct 2, 2013

### PhysicsGente

That makes sense. I also realized that it can be integrated by parts using the same reasoning of vanishing at the boundaries. Thanks!